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Over [174]
4 years ago
12

In a piston, the volume was noticed to be 2.50 liters at 25°C. If the temperature rose to 55°C, what is the new volume?

Chemistry
2 answers:
GrogVix [38]4 years ago
8 0

Answer:

The new volume is 2.75L

Explanation:

This is another case of Charles Gay Lussac law, where the volume varies directly proportional to T°, if the pressure keeps on constant.

V₁ / T₁  =  V₂ / T₂

25°C + 273 = 298K

55°C + 273 = 328K

2.5L / 298K = V₂ / 328K

(2.5L / 298K) . 328K = V₂

2.75L = V₂

saul85 [17]4 years ago
6 0

Answer:

The new volume is 2.75 L

Explanation:

Step 1: Data given

Initial volume = 2.50 L

Initial temperature = 25 °C = 298.15 K

Temperature rises to 55°C

Step 2: Calculate the new volume

V1 / T1 = V2 / T2

⇒ with V1 = The initial volume = 2.50 L

⇒ with T1 = The initial temperature = 298.15 Kelvin

⇒ with V2 = the new volume = TO BE DETERMINED

⇒ with T2 = The new temperature = 55 °C = 328.15 K

V2 = (V1/T1)*T2

V2 = (2.50/ 298.15)*328.15

V2 = 2.75 L

The new volume is 2.75 L

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